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RVUNL’21 EE: Daily Practices Quiz 05-Aug-2021

Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06
Time: 06 min

Q1. Which among the following is termed as the drift velocity of the charge carrier per unit electric field?
(a) Resistivity
(b) Current density
(c) Mobility
(d) Relative permittivity

Q2. The length of a copper wire is increased by 2%. Calculate the percentage change (in %) in its resistance.
(a) 2
(b) 3
(c) 4
(d) 5

Q3. In the salient pole of 3-phase synchronous machine the rotor will have
(a) Larger diameter, shorter axial length
(b) Smaller diameter, larger axial length
(c) Larger diameter, larger axial length
(d) Smaller diameter, shorter axial length

Q4. A 3-phase, 6-pole, 50 Hz induction motor has a slip of 1 % at no load and 3 % at full load. The frequency of rotor current at standstill will be:
(a) 50 Hz
(b) 100 Hz
(c) 1.5 Hz
(d) Zero

Q5. A 500 KVA transformer has constant loss of 500 W and copper losses at full load are 2000 W. Then at what load, is the efficiency maximum?
(a) 1000 KVA
(b) 500 KVA
(c) 250 KVA
(d) 125 KVA

Q6. System is said to be marginally stable, if
(a) Gain crossover frequency > Phase crossover frequency
(b) Gain crossover frequency = Phase crossover frequency
(c) Gain crossover frequency < Phase crossover frequency
(d) Gain crossover frequency ≠ Phase crossover frequency

SOLUTIONS
S1. Ans.(c)
Sol. drift velocity (V_d )= mE
⇒▭(m=V_d/E)
Where;
m=mobility
V_d=drift velocity
E=electric field

S2. Ans.(c)
Sol. R=ρl/A
∆ R=∆l+∆A
=2+2=4%
Method- II, if l is increased by n%
Then R_new=n^2×R_old
R^1= (2)^2×R
=4R
∵ New resistance will be increased by 4% of previous value.

S3. Ans.(a)
Sol. 3-phase synchronous machine rotor construction:
• Salient pole rotor: larger diameter, shorter axial length
• Cylindrical pole rotor: smaller diameter, larger axial length

S4. Ans.(a)
Sol. f_r=sf
At standstill condition: slip(s)=1
So, frequency of rotor current at standstill=supply frequency=50 Hz.

S5. Ans.(c)
Sol. The load at which maximum efficiency occurs:
η_max=KVA×√(P_i/P_(cu(fl)) )=500×√(500/2000)=250 KVA

S6. Ans.(b)
Sol.
ω_pc>ω_gc ⇒stable system
ω_pc<ω_gc ⇒unstable system
ω_pc=ω_gc ⇒marginally stable

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